Animated Solution for Mathematics - Probability: Let $S=\{1,2,3,\dots ,2022\}$. Then the probability, that a randomly chosen number $n$ from the set $S$ such that $HCF(n,2022)=1$, is:

Imagine you are standing before a vast sequence of numbers, from $1$ to $2022$. You are asked to pick one at random, and you need to find the probability that your chosen number $n$ is 'co-prime' to $2022$.

This means $HCF(n,2022)=1$. It sounds like a daunting task, but let's break it down together.

First, let's look at our playground. We have a set $S=\{1,2,3,\dots ,2022\}$.

If we pick a number at random, our total number of possible outcomes is simply $N=2022$. This is our denominator.

We need to count how many numbers $n$ in this set satisfy $HCF(n,2022)=1$. In the world of number theory, this is where Euler's Totient Function, denoted by $\varphi (N)$, becomes our most powerful weapon.

It is specifically designed to count the number of integers up to $N$ that are co-prime to $N$. The formula is:

Here, $p_1,p_2,\dots ,p_k$ are the distinct prime factors of $N$.

To use this formula, we must first uncover the prime building blocks of $2022$. Let's be detectives.

$2022$ is clearly even, so $2022=2\times 1011$. Now, look at $1011$. The sum of its digits is $1+0+1+1=3$, which is divisible by $3$. So, $1011=3\times 337$.

Now, is $337$ prime? If we check primes up to $\sqrt{337}\approx 18$, we find it doesn't divide by $2,3,5,7,11,13,$ or $17$. Yes, $337$ is prime! Our distinct prime factors are $2,3,$ and $337$.

Now, we plug these into our formula:

Let's simplify the brackets: $(1-\frac{1}{2})=\frac{1}{2}$, $(1-\frac{1}{3})=\frac{2}{3}$, and $(1-\frac{1}{337})=\frac{336}{337}$. The expression becomes:

Notice the beauty here? The denominator $2\times 3\times 337$ is exactly $2022$. They cancel out perfectly! We are left with $1\times 2\times 336=672$. There are exactly $672$ favorable outcomes.

Finally, the probability $P$ is the ratio of favorable outcomes to total outcomes:

To simplify, we divide both by $6$, giving us $P=\frac{112}{337}$.

And there it is! A complex counting problem solved with the elegance of number theory. Remember, whenever you see 'co-prime' in a JEE problem, let Euler's Totient Function be your first thought.